Question: $w_1=6[\cos(12^\circ)+i\sin(12^\circ)]$ $w_2=15[\cos(18^\circ)+i\sin(18^\circ)]$ Express the product of $w_1$ and $w_2$ in polar form. Angle measure should be in degrees. No need to use the degree symbol. $w_1\cdot w_2= $
Answer: Background For any two complex numbers $z_1$ and $z_2$ (whose radii are $r_1$ and $r_2$ and angles are $\theta_1$ and $\theta_2$ ): The radius of $z_1\cdot z_2$ is the product of the original radii, $r_1 \cdot r_2$. The angle of $z_1\cdot z_2$ is the sum of the original angles, $\theta_1 + \theta_2$. In other words, suppose the polar forms of $z_1$ and $z_2$ are as follows, $z_1 = r_1[\cos(\theta_1) + {i}\sin(\theta_1)]$ $z_2 = r_2[\cos(\theta_2) + {i}\sin(\theta_2)]$, then the polar form of their product is: $z_1z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + {i}\sin(\theta_1 + \theta_2)]$. [How do we get this?] Finding the radius of $w_1\cdot w_2$ $w_1=6[\cos(12^\circ)+i\sin(12^\circ)]$ $w_2=15[\cos(18^\circ)+i\sin(18^\circ)]$ Here, $r_1=6$ and $r_2=15$. Therefore, the radius of $w_1\cdot w_2$ is $r_1\cdot r_2=6\cdot15=90$. Finding the angle of $w_1\cdot w_2$ $w_1=6[\cos(12^\circ)+i\sin(12^\circ)]$ $w_2=15[\cos(18^\circ)+i\sin(18^\circ)]$ Here, $\theta_1=12^\circ$ and $\theta_2=18^\circ$. Therefore, the angle of $w_1\cdot w_2$ is $\theta_1+\theta_2=12^\circ+18^\circ=30^\circ$. Summary We found that the radius of $w_1\cdot w_2$ is $90$ and its angle is $30^\circ$. Therefore, $w_1\cdot w_2=90\left(\cos(30^\circ)+i\sin(30^\circ)\right)$